Conlusion: One way to check our factoring is to graph the given polynomial and check that the x intercepts corresponds to the zeros of the factors included in the factorization.
Question c): Factor 15 x 2 - 3 x + 10 x - 2

Solution

c) Find a common factor in 15 x 2 - 3 x and factor it as follows:

15 x 2 - 3 x = 3 x (5 x - 1)

We next find a common factor in the 10 x - 2 and factor it as follows:

10 x - 2 = 2 (5 x - 1)

Use the common factor (5 x - 1) to factor the given polynomial as follows:

15 x 2 - 3 x + 10 x - 2 = ( 15 x 2 - 3 x ) + (10 x - 2)

= 3 x (5 x - 1) + 2 (5 x - 1) = (5 x - 1)(3 x + 2)

Question d): Factor 4 x 2 + x - 3

Solution

d) The given polynomial has three terms with no common factor. One way to factor is to rewrite it replacing x by 4 x - 3 x as follows:

4 x 2 + x - 3 = 4 x 2 + 4 x - 3 x - 3

We can now factor 4 x 2 + 4 x as follows:

4 x 2 + 4 x = 4 x (x + 1)

We next factor - 3 x - 3 as follows:

- 3 x - 3 = - 3 (x + 1)

Use the common factor (x + 1) to factor the given polynomial as follows:

4 x 2 + x - 3 = 4 x 2 + 4 x - 3 x - 3 = (4 x 2 + 4 x) + (- 3 x - 3)

= 4 x (x + 1) - 3 (x + 1) = (x + 1)(4 x - 3)

Question e): Factor x 2 y + 3 x + x 2 y 2 + 3 x y

Solution

e) Note that x is a common factor to all terms in the given polynomial. Hence we start by factoring as follows:

x 2 y + 3 x + x 2 y 2 + 3 x y = x( x y + 3 + x y 2 + 3 y)

Rewrite by grouping terms as follows

x 2 y + 3 x + x 2 y 2 + 3 x y = x( (x y + x y 2) + (3 + 3 y) )

The terms in (x y + x y 2) has the factor x y and the terms in (3 + 3 y) has the common factor 3. Hence we factor as follows

x 2 y + 3 x + x 2 y 2 + 3 x y = x( (x y + x y 2) + (3 + 3 y) )

= x( x y (1 + y) + 3 (1 + y) ) = x (1 + y)( x y + 3)

Question f): Factor 3 x 2 + 3 x y - x + 2 y - 2

Solution

f) Note that there are 5 terms in the given polynomial with common factor to all of them. Rewrite the polynomial replacing - x by - 3 x + 2 x as follows.